As, block has a tendency to slide downwards, friction acts in upward direction.
Box will not slide if friction balances component of weight acting in downward direction.
$\Rightarrow \quad f \geq m g \sin \theta$
Here, $f=$ friction $=\mu N$
or $\quad f=\mu[(m g \cos \theta)+F]$
As, applied force is perpendicular to surface of incline.
Normal reaction, $N=$ net perpendicular to surface force
$$
=m g \cos \theta+F
$$

Hence, by Eqs. (i) and (ii)we have
$$
\begin{aligned}
& \mu(m g \cos \theta)+\mu F \geq m g \sin \theta \\
& \text { or } F \geq \frac{m g \sin \theta-\mu m g \cos \theta}{\mu}
\end{aligned}
$$
here,
$$
\begin{aligned}
& m=2 \mathrm{~kg} \\
& g=10 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
$$
$$
\begin{gathered}
\sin \theta=\sin 30^{\circ}=\frac{1}{2} \\
\mu=0.2
\end{gathered}
$$

So, $F \geq\left(\frac{2 \times 10 \times \frac{1}{2}-0.2 \times 2 \times 10 \times \frac{\sqrt{3}}{2}}{0.2}\right)$
$$
\begin{aligned}
& \Rightarrow \quad F \geq 50-17.32 \\
& \Rightarrow \quad F \geq 3267
\end{aligned}
$$
or $F \geq 327$
Hence, minimum value of $F$ is $32.7 \mathrm{~N}$.