$$ r = R + \frac{R}{3} = \frac{4R}{3} $$

For a circular orbit, the gravitational force provides the required centripetal force:

$$ \frac{GMm}{r^2} = \frac{mv^2}{r} $$

which simplifies to

$$ v = \sqrt{\frac{GM}{r}}. $$

The angular momentum $$L$$ of the satellite is given by

$$ L = mvr. $$

Substitute the values:

Satellite mass: $$ m = \frac{M}{2} $$

Orbital radius: $$ r = \frac{4R}{3} $$

Speed: $$ v = \sqrt{\frac{GM}{\frac{4R}{3}}} = \sqrt{\frac{3GM}{4R}} $$

Thus,

$$ L = \frac{M}{2} \cdot \frac{4R}{3} \cdot \sqrt{\frac{3GM}{4R}}. $$

Simplify the expression:

$$ L = \frac{M \cdot 4R}{6} \sqrt{\frac{3GM}{4R}} = \frac{2MR}{3} \sqrt{\frac{3GM}{4R}}. $$

Notice that the square root can be combined as:

$$ \sqrt{\frac{3GM}{4R}} = \sqrt{\frac{3}{4}} \sqrt{\frac{GM}{R}}. $$

Therefore,

$$ L = \frac{2MR}{3} \cdot \sqrt{\frac{3}{4}} \sqrt{\frac{GM}{R}} = \frac{2MR \sqrt{3}}{3 \cdot 2} \sqrt{\frac{GM}{R}} = \frac{M \sqrt{3}}{3} \sqrt{GMR}, $$

where we used the fact that $$ R \sqrt{\frac{1}{R}} = \sqrt{R} $$ to form $$ \sqrt{GMR} $$.

So the final expression is:

$$ L = \frac{M}{\sqrt{3}} \sqrt{GMR}. $$

It is given that the angular momentum can also be expressed as:

$$ L = M \sqrt{\frac{GMR}{x}}. $$

Equate the two expressions:

$$ \frac{M}{\sqrt{3}} \sqrt{GMR} = M \sqrt{\frac{GMR}{x}}. $$

Cancel $$M$$ and $$\sqrt{GMR}$$ on both sides (assuming they are nonzero):

$$ \frac{1}{\sqrt{3}} = \sqrt{\frac{1}{x}}. $$

Taking squares on both sides:

$$ \frac{1}{3} = \frac{1}{x}. $$

Thus,

$$ x = 3. $$