Let the locus of the centre (α, β), β > 0 , of the circle which...

Since the circle with centre \left(α, β\right) touches the circle ${x}^{2} + {\left(y - 1\right)}^{2} = 1$ externally and also touches $x - \text{axis}$,

therefore the distance between the centres will be equal to the sum of the radius of both circles

i.e. \sqrt{{\left(α - 0\right)}^{2} + {\left(β - 1\right)}^{2}} = β + 1

\to {α}^{2} = 4β

Hence the locus of $L$ will be ${x}^{2} = 4y$

Now, the area bounded by ${x}^{2} = 4y \text{ and } y = 4$ will be

A = 2{∫}_{0}^{4}\left(4 - \dfrac{{x}^{2}}{4}\right)dx

$= 2{\left(4x - \dfrac{{x}^{3}}{12}\right)}_{0}^{4} = \dfrac{64}{3}$