The figure given below, shows the downward motion of a body on the inclined plane, then


The equation of acceleration in downwards motion of the block,
$$
a=g \sin \theta-\mu g \cos \theta
$$
Given, mass of a block, $m=10 \mathrm{~kg}, \theta=45^{\circ}, \mu=0.3$ and time, $t=2 \mathrm{~s}$, Putting the given values in Eq. (i), we get
$$
\begin{aligned}
\Rightarrow \quad a & =g\left(\sin 45^{\circ}-0.3 \cos 45^{\circ}\right) \\
& =\frac{10}{\sqrt{2}}(1-0.3)=\frac{10 \times 0.7}{\sqrt{2}} \mathrm{~m} / \mathrm{s}^2
\end{aligned}
$$
Displacement of the block, from the second equation of the motion,
$$
\begin{array}{cc}
s=u t+\frac{1}{2} a t^2 & (\because u=0) \\
s=\frac{1}{2} \times \frac{10 \times 0.7}{\sqrt{2}} \times(2)^2=\frac{7 \times 4}{2 \sqrt{2}}=7 \sqrt{2} \mathrm{~m} &
\end{array}
$$
So, the correct option is (1).