Interpret the setup

Three identical blocks $A, B, C$ each have mass $m=2\,\text{kg}$ They move together on a frictionless surface with acceleration $a=0.6\,\text{m s}^{-2}$ under an applied force $F=10.2\,\text{N}.$

We need the tension in the string between blocks $B$ and $C$.

Find the net force needed for acceleration

Total mass of the three blocks is $M=3m=3\times 2=6\,\text{kg}.$

Hence the net force accelerating the system is $F_{\text{net}}=Ma=6\times 0.6=3.6\,\text{N}.$

Find the total resisting force

Since the applied force is $10.2\,\text{N}$ but the system accelerates as if net force is only $3.6\,\text{N}$, there must be an opposing force of magnitude $R=10.2-3.6=6.6\,\text{N}.$

For three identical blocks, this resisting force is shared equally: $\text{resistance on each block} = \frac{6.6}{3}=2.2\,\text{N}.$

Consider block $C$ alone

On block $C$, the horizontal forces are:

• tension $T$ pulling forward,
• resistance $2.2\,\text{N}$ backward.

Using Newton's second law: $T-2.2 = ma$ $T-2.2 = 2\times 0.6 = 1.2$ $T = 1.2+2.2 = 3.4\,\text{N}.$

Answer

Therefore, the tension in the string between $B$ and $C$ is $\boxed{3.4\,\text{N}}$

So the correct option is B.