The outward pull on a metal plate of an area $0.01m^2$ having a charge density of $50 μC/m^2$ is

1.4 N

$\text{Electric Field due to rod is at its surface } E = \frac{\sigma}{2\epsilon_0} = \frac{50\times 10^{-6}}{2\times 8.85\times 10^{-12}} = 2.82 \times 10^6 N/C$

$ F = qE = \sigma A E = 50\times 10^{-6} \times 10^{-2} \times 2.82 \times 10^6 = 1.4N$