Let us assume that the two blocks move together, without slipping, relative to each other. The acceleration of the system in that case is

$a=\frac{F\cos 60°}{m_1+m_2}$

$a=\frac{150\times {\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{20+30}=1.5 \mathrm{m} {\mathrm{s}}^{-2}$

In this case, if the frictional force acting between the two blocks is $f$, then writing the Newton's second law of motion, for the block of mass $m_{\mathrm{1,}}$ we get

$T-f=m_{1}a$

$150-f=20\times 1.5=30$

$f=120 \mathrm{N}$

$f_{\max }=0.6\times 200=120 \mathrm{N}$

Since $f\le f_{\max }$, our assumption about the two blocks moving together is correct and hence the acceleration of the blocks is $1.5 \mathrm{m} {\mathrm{s}}^{-2}$