The general term in the expansion of the binomial expression $(a+b)^n$ is

$$ T_{r+1}={ }^n C_r a^{n-r} b^r $$

Therefore, the general term in the expansion of the binomial expression

$x^2\left(\sqrt{x}+\frac{\lambda}{x^2}\right)^{10}$ is

$$ \begin{aligned} T_{r+1} & =x^2\left({ }^{10} C_r(\sqrt{x})^{10-r}\left(\frac{\lambda}{x^2}\right)^r\right) \\\\ & ={ }^{10} C_r x^2 \cdot x^{\frac{10-r}{2}} \lambda^r x^{-2 r} \\\\ & ={ }^{10} C_r \lambda^r x^{2+\frac{10-r}{2}-2 r} \end{aligned} $$

Now, for the coefficient of $x^2$,

$$ \begin{aligned} 2+\frac{10-r}{2}-2 r =2 \\\\ \Rightarrow \frac{10-r}{2}-2 r =0 \\\\ \Rightarrow 10-r =4 r \Rightarrow r=2 \end{aligned} $$

So, the coefficient of $x^2$ is ${ }^{10} C_2 \lambda^2=720$

$$ \begin{aligned} \Rightarrow & \frac{10 !}{2 ! 8 !} \lambda^2 =720 \\\\ \Rightarrow & \frac{10 \cdot 9 \cdot 8 !}{2 \cdot 8 !} \lambda^2 =720 \\\\ \Rightarrow & 45 \lambda^2 =720 \\\\ \Rightarrow & \lambda^2 =16 \\\\ \Rightarrow & \lambda = \pm 4 \end{aligned} $$

Given the problem asks for the positive value of $\lambda$, $\lambda = 4$.

So, the correct option is (A) 4.