= Solution
Let $\zeta=e^{2\pi i/7}$ and let $\lambda_k(a)=\zeta^k$. Conjugation by $b$ partitions the six nontrivial characters of $H=\langle a\rangle$ into the two orbits
$$
\{1,2,4\},
\qquad
\{3,5,6\}.
$$
Inducing one representative from each orbit gives two irreducible degree-three characters $\psi,\overline\psi$. On the five classes represented by $1,a,a^3,b,b^2$ their values are
$$
\psi=(3,\zeta+\zeta^2+\zeta^4,
\zeta^3+\zeta^5+\zeta^6,0,0),
$$
with the two cyclotomic sums interchanged for $\overline\psi$. Their squared degrees, together with the three linear characters, sum to
$$
3\cdot1^2+2\cdot3^2=21,
$$
so these are all irreducible characters.
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