If $a$ is the acceleration produced in the block, then
$ma = mg\sin\theta -f_k$
(where $f_k $ is force of kinetic friction)
$= mg\sin0 -μ_k N$                   $(as\ \ f_k. = μ_k N)$
$= mg\sin0 - μ_kmg\cos\theta$            $(as\ \  N = mgcos\theta)$
$a = g(\sin\theta - μ_k\cos\theta)$
As $ g = 10 ms^{- 2}$ and $0 = 30°$
$a = ( 10 ms^{- 2})(sin30^o -μ_kcos30^o)$  ..........$(i)$

If $s$ is the distance travelled by the block in time $t$, then

$S=\frac{1}{2} at^2$                $(u=0)$

$a=\frac{2s}{t^2}$

But $s=4.0 \ m$ and $t=4.0 \ s$ (given)

$a=\frac{2(4.0\ m)}{(4.0\ s)^2}=\frac{1}{2}\ ms^{-2}$

Substituting this value of $a$ in eqn. $(i)$, we get

$\frac{1}{2}\ ms^{-2} =10\ ms^{-2}\left( \frac{1}{2}-\mu_k \frac{\sqrt3}{2} \right)$

$\frac{1}{10}=1-\sqrt3 \mu_k$ or $\sqrt3 \mu_k=1-\frac{1}{10}=\frac{9}{10}=0.9$

$\mu_k =\frac{0.9}{\sqrt3}=0.5$