1. Find the tension in wire $B$

Since the system is in equilibrium and friction is neglected, wire $B$ supports only block $R$.

Mass of each block $= 3\,\text{kg}$, so weight of each block is $mg = 3 \times 10 = 30\,\text{N}.$

Hence, $T_B = 30\,\text{N}.$

2. Use the relation for longitudinal strain

Longitudinal strain is $\text{strain} = \frac{\text{stress}}{Y} = \frac{T/A}{Y} = \frac{T}{AY}.$

Given:

• $A = 0.005\,\text{cm}^2$
• $Y = 2 \times 10^{11}\,\text{N m}^{-2}$

Convert area into SI units: $0.005\,\text{cm}^2 = 0.005 \times 10^{-4}\,\text{m}^2 = 5 \times 10^{-7}\,\text{m}^2.$

Therefore, $\text{strain in } B = \frac{30}{(5 \times 10^{-7})(2 \times 10^{11})}.$

Now, $(5 \times 10^{-7})(2 \times 10^{11}) = 10 \times 10^4 = 10^5.$

So, $\text{strain} = \frac{30}{10^5} = 3 \times 10^{-4}.$

3. Final answer

The longitudinal strain on wire $B$ is $3 \times 10^{-4}.$

So the required integer is: $\boxed{3}$