According to question a $F$  force act on the block 

Now F.B.D. for block $3 \mathrm{kg}$ 

Where, $N$ is normal provided by  $2 \mathrm{kg}$ block 
$f$ is force of friction 
$\mu$ if coefficient of friction 
According to newton's second law
$\sum _{}^{}F=ma$
Applied in the direction of incline
$3g \sin \mathrm{\theta }+N-f=3a$
$f=\mu N_1$
Here $N_1$ is normal force applied by the incline on $\text{2 Kg}$ block
$3g \sin \mathrm{\theta }+N-3\mu g \cos \mathrm{\theta }=3a ...\left(1\right)$

 for $2 \mathrm{kg}$ block

$\therefore F+2g \sin \mathrm{\theta }- N-2\mu g \cos \mathrm{\theta }=2a ...\left(2\right)$
 adding$\left(1\right)$  and $\left(2\right)$
We get 
$5g \sin \mathrm{\theta }-5\mu g \cos \mathrm{\theta }+F=5a$ 
given  $\theta = 37°$ 
$F=20 \mathrm{N}$ 
$\mu =0.1$ 
$\Rightarrow 5 \times 10 \times \left(\frac{3}{5}\right) - 50(0.1)\left(\frac{4}{5}\right) + 20 = 5a$ $\Rightarrow a=\frac{46}{5} \mathrm{m} {\mathrm{s}}^{-2}$ 
Now putting the value of acceleration in $\left(2\right)$ 
We get 
$20+20\left(\frac{3}{5}\right)-N-\left(2\times \frac{4}{5}\right)=2\times \frac{46}{5}$ $20+12-N-\left(\frac{8}{5}\right)=\frac{92}{5}$ 
$\Rightarrow N=12 \mathrm{N}$