From the given figure we can see that 

$\sin {\theta }_0=\frac{A}{l}=\frac{10}{100}=\frac{1}{10}$

From conservation of energy $\frac{m{v}^2}{2}=mgl(1-\cos {\theta }_0)$

Maximum tension occurs at mean position

$$\begin{gathered} T=mg+\frac{m{v}^2}{l} \\ \Rightarrow T=mg+2mg(1-\cos {\theta }_0) \end{gathered}$$

Substituting the values,

$$\begin{gathered} T=mg(1+2(1-\sqrt{1-{\sin }^2{\theta }_0}) \\ \Rightarrow T=mg(3-2\sqrt{1-\frac{1}{100}}) \\ \Rightarrow T=0.25\times 9.8(3-2(1-\frac{1}{200})) \\ \Rightarrow T=0.25\times 9.8\times 1.01=2.4745 \mathrm{N} \end{gathered}$$

It is given that

 $$\begin{gathered} T=\frac{x}{40} \\ \Rightarrow x=40\times 2.4745=98.98\approx 99 \end{gathered}$$