Given potential energy

$U(x)=\frac{a}{x^{12}}-\frac{b}{x^6}$

The dissociation energy is defined as

$D=U(\infty)-U_{\text{equilibrium}}$

Find the equilibrium separation

At equilibrium, potential energy is minimum, so

$\frac{dU}{dx}=0$

Differentiate:

$\frac{dU}{dx}=-12a x^{-13}+6b x^{-7}$

Set equal to zero:

$-12a x^{-13}+6b x^{-7}=0$

$6b x^{-7}=12a x^{-13}$

Multiply by $x^{13}$:

$6b x^6=12a$

$x^6=\frac{2a}{b}$

Compute the potential energy at equilibrium

Using

$U=\frac{a}{x^{12}}-\frac{b}{x^6}$

and

$x^6=\frac{2a}{b} \implies x^{12}=\left(\frac{2a}{b}\right)^2=\frac{4a^2}{b^2}$

Therefore,

$\frac{a}{x^{12}}=a\cdot \frac{b^2}{4a^2}=\frac{b^2}{4a}$

and

$\frac{b}{x^6}=b\cdot \frac{b}{2a}=\frac{b^2}{2a}$

So,

$U_{\text{eq}}=\frac{b^2}{4a}-\frac{b^2}{2a}=-\frac{b^2}{4a}$

Compute $U(\infty)$

As $x\to \infty$,

$\frac{a}{x^{12}}\to 0, \qquad \frac{b}{x^6}\to 0$

hence,

$U(\infty)=0$

Dissociation energy

$D=U(\infty)-U_{\text{eq}}=0-\left(-\frac{b^2}{4a}\right)=\frac{b^2}{4a}$

Match with options

$D=\frac{b^2}{4a}$

So the correct option is C.