= Solution
Write $N=\langle a\rangle$. The relation $bab^{-1}=a^2$ lets every element be expressed as $a^ib^j$ with $0\leq i<7$ and $0\leq j<3$. There are 21 such expressions, so the assumption $|G|=21$ makes them distinct. In particular $a$ has order seven, $b$ has order three, and $N$ is a <normal subgroup>. This is the <nonabelian group of order 21>, the <semidirect product> $C_7\rtimes C_3$ with the action given by multiplication by two modulo seven.
Conjugation by $a$ fixes $N$, while conjugation by $b$ sends $a^i$ to $a^{2i}$. The nonidentity <conjugacy classes> inside $N$ are therefore
$$
K_1=\{a,a^2,a^4\},\qquad
K_2=\{a^3,a^5,a^6\}.
$$
For the other elements, direct calculation gives
$$
a^t(a^ib^j)a^{-t}=a^{i+(1-2^j)t}b^j.
$$
When $j=1$ or $2$, the coefficient $1-2^j$ is nonzero modulo seven, so letting $t$ vary gives every member of the <coset> $Nb^j$. These <cosets> cannot merge under <conjugation>, since $G/N$ is an <abelian group> and conjugation acts trivially on that <quotient group>. Thus the five <conjugacy classes> have representatives and sizes
$$
\begin{array}{c|ccccc}
\text{representative}&1&a&a^3&b&b^2\\\hline
\text{class size}&1&3&3&7&7
\end{array}.
$$
To find the <linear characters>, note that the <commutator subgroup> $G'$ contains $bab^{-1}a^{-1}=a$. The <quotient group> $G/N\cong C_3$ is abelian, so $G'\subseteq N$; hence $G'=N$. The <abelianization> is $C_3$. With $\omega=e^{2\pi i/3}$, its three <linear characters> are
$$
\lambda_j(a)=1,\qquad\lambda_j(b)=\omega^j,\qquad j=0,1,2.
$$
This exhausts the <linear characters> because <one-dimensional characters factor through the abelianization>.
For the remaining <irreducible characters>, put $\zeta=e^{2\pi i/7}$ and take the <linear characters> $\theta_t$ of $N$ defined by $\theta_t(a)=\zeta^t$. The six nontrivial <linear characters> form two orbits under conjugation, with exponent sets $\{1,2,4\}$ and $\{3,5,6\}$. Define the <induced characters>
$$
\psi=\operatorname{Ind}_N^G\theta_1,\qquad
\overline\psi=\operatorname{Ind}_N^G\theta_3.
$$
For $a^i\in N$, the <character formula for an induced representation> gives
$$
\psi(a^i)=\zeta^i+\zeta^{2i}+\zeta^{4i},\qquad
\overline\psi(a^i)=\zeta^{3i}+\zeta^{5i}+\zeta^{6i}.
$$
Both vanish outside $N$, since no element outside a <normal subgroup> can be conjugate into it. This is <induction from an abelian normal subgroup with a free character orbit>. For completeness, its irreducibility here follows directly from <Frobenius reciprocity> and <character orthogonality> on the <cyclic group> $N$:
$$
\langle\psi,\psi\rangle_G
=\langle\theta_1,\operatorname{Res}_N^G\psi\rangle_N
=\langle\theta_1,\theta_1+\theta_2+\theta_4\rangle_N=1.
$$
The same calculation with $\theta_3,\theta_5,\theta_6$ proves that $\overline\psi$ is an <irreducible character>.
Let
$$
\alpha=\zeta+\zeta^2+\zeta^4,\qquad
\beta=\zeta^3+\zeta^5+\zeta^6=\overline\alpha.
$$
The sum of the nontrivial seventh <roots of unity> gives $\alpha+\beta=-1$. Squaring and reducing exponents modulo seven gives
$$
\alpha^2=\alpha+2\beta=-\alpha-2.
$$
Thus $\alpha,\beta$ are the two <polynomial roots> of $T^2+T+2$, and
$$
\alpha=\frac{-1+i\sqrt7}{2},\qquad
\beta=\frac{-1-i\sqrt7}{2}.
$$
To fix the signs with our specified $\zeta$, observe that $\operatorname{Im}\alpha=\sin(2\pi/7)+\sin(3\pi/7)-\sin(\pi/7)>0$, using the increase of sine on $(0,\pi/2)$. These <Gaussian periods for seventh roots of unity> make the answer explicit.
The resulting <character table> is
$$
\boxed{\begin{array}{c|ccccc}
&1&a&a^3&b&b^2\\\hline
\text{class size}&1&3&3&7&7\\\hline
\lambda_0&1&1&1&1&1\\
\lambda_1&1&1&1&\omega&\omega^2\\
\lambda_2&1&1&1&\omega^2&\omega\\
\psi&3&\alpha&\beta&0&0\\
\overline\psi&3&\beta&\alpha&0&0
\end{array}.}
$$
The two degree-three <irreducible characters> are distinct because $\alpha\ne\beta$. Together with the three <linear characters>, they exhaust the <irreducible characters>: the <sum of squares of irreducible degrees> is $3\cdot1^2+2\cdot3^2=21=|G|$. Equivalently, we have five distinct <irreducible characters> for the five <conjugacy classes>.
As a check on part (b), inversion interchanges $K_1$ and $K_2$, and also interchanges the two <conjugacy classes> outside $N$. Only the identity <conjugacy class> is real. Correspondingly $\lambda_0$ is the only <real-valued character> among the <irreducible characters>; $\lambda_1,\lambda_2$ and $\psi,\overline\psi$ are two conjugate pairs.
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