The box of a pin hole camera, of length L, has a hole of radius a....

In diffraction, first minima, we have

$sin\theta = \dfrac{\lambda}{a}$

$b = a + \dfrac{L\lambda}{a} . . . \left(i\right)$

Then, minimum size of a spot, we get

$\dfrac{∂ b}{∂ a} = 0 ⇒ 1 - \dfrac{L\lambda}{{a}^{2}} = 0 ⇒ a = \sqrt{\lambdaL} . . . \left(ii\right)$

So,

{b}_\min = \sqrt{\lambdaL} + \sqrt{\lambdaL}

$= 2\sqrt{\lambdaL}$$= \sqrt{4\lambdaL}$