We have,

\(f _1=20\,N , f _2=10\,N\)

\(m_1=2\,kg , m_2=4\,kg\)

Let tension between the blocks is \(T\)

Here, \(F_1 > F_2\)

So, the net force acts in the right direction.

for block of mass \(2 kg\),

\(T - F _2=2 a\)

\(T-10=2 a \ldots \ldots \ldots(1)\)

For the block of mass \(4\,kg\),

\(F_1-T=4 a\)

\(20-T=4 a .\)

BY solving the above two equation the we get,

\(10=6 a\)

\(\Rightarrow a=\frac{5}{3} m / s ^2\)

Thus,

From equation \((1)\) we get,

\(T=10+\frac{10}{3}\)

\(=\frac{40}{3} \,N\)