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Write $N=⟨a⟩$. The relation $ba{b}^{-1}=a^2$ lets every element be expressed as $a^i{b}^j$ with $0\le i<7$ and $0\le j<3$. 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 $C_7⋊C_3$ 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 thereforeFor the other elements, direct calculation givesWhen $j=1$ or $2$, the coefficient $1-2^j$ is nonzero modulo seven, so letting $t$ vary gives every member of the coset $N{b}^j$. 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

To find the linear characters, note that the commutator subgroup $G^{\prime }$ contains $ba{b}^{-1}{a}^{-1}=a$. The quotient group $G/N\cong C_3$ is abelian, so $G^{\prime }\subseteq N$; hence $G^{\prime }=N$. The abelianization is $C_3$. With $\omega =e^{2\pi i/3}$, its three linear characters areThis 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 charactersFor $a^i\in N$, the character formula for an induced representation givesBoth 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 $N$:The same calculation with ${\theta }_3,{\theta }_5,{\theta }_6$ proves that $\overline{\psi }$ is an irreducible character.

LetThe sum of the nontrivial seventh roots of unity gives $\alpha +\beta =-1$. Squaring and reducing exponents modulo seven givesThus $\alpha ,\beta$ are the two polynomial roots of $T^2+T+2$, andTo fix the signs with our specified $\zeta$, observe that $\mathrm{Im}\alpha =\sin (2\pi /7)+\sin (3\pi /7)-\sin (\pi /7)>0$, 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 isThe 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 $3\cdot 1^2+2\cdot 3^2=21=|G|$. 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.