For an ideal solenoid $B_{\text{out}} $ = 0 and $B_{\text{in}} = \frac{\mu_0}{4 \pi} . 4 \pi \frac{Ni}{l}$
$=\frac{4 \pi \times 10^{-7} \times 300 \times 2}{0.7} = 1.076 \times 10^{-3} \,T$.
And $\phi = BA \,cos \,\theta$ or $\phi = \phi_{\text{in}} + \phi_{\text{out}}$
$= B_{\text{in}} \times (\pi r^2) \times \, cos\, 60\,{\circ} + 0$ $( \because \, B_{\text{out}} = 0)$
$= 1.076 \times 10^{-3}) \times ( \pi \times 4 \times 10^{-4}) \times \frac{1}{2}$
$= 13.52 \times 10^{-7} \times \frac{1}{2} = 6.76 \times 10^{-7}\, Wb$