According to the empirical formula relating the radius of a nucleus ($ R $) with its mass number ($ A $), we know that the radius of a nucleus is proportional to the cube root of its mass number. This relationship is given as:

$$ R = R_0 A^{1/3} $$

where $ R_0 $ is a constant with an approximate value of 1.2 fermis.

Given that for a nucleus with a mass number 64 has a radius of 4.8 fermis, we can write:

$$ 4.8 \text{ fermi} = R_0 \times 64^{1/3} $$

Now another nucleus has a radius of 4 fermis:

$$ 4 \text{ fermi} = R_0 \times A'^{1/3} $$

Where $ A' $ is the mass number of the other nucleus.

Let's solve for $ R_0 $ from the first equation:

$$ R_0 = \frac{4.8 \text{ fermi}}{64^{1/3}} $$

Now we're going to find the mass number $ A' $ using the second equation and substituting $ R_0 $ from the above:

$$ 4 \text{ fermi} = \left( \frac{4.8 \text{ fermi}}{64^{1/3}} \right) \times A'^{1/3} $$

Now, we want to find $ A' $ in terms of $ x $ as given by the equation in the question:

$$ A' = \frac{1000}{x} $$

Substitute $ A' $ in the equation above, we get:

$$ 4 = \left( \frac{4.8}{64^{1/3}} \right) \times \left( \frac{1000}{x} \right)^{1/3} $$

Let's solve for $ x $:

$$ (4)^3 = \left( \frac{4.8}{64^{1/3}} \right)^3 \times \frac{1000}{x} $$

$$ 4^3 \times x = \left( \frac{4.8}{64^{1/3}} \right)^3 \times 1000 $$

$$ x = \left( \frac{\left( \frac{4.8}{64^{1/3}} \right)^3 \times 1000}{4^3} \right) $$

Now calculate the values:

$$ x = \left( \frac{\left( \frac{4.8}{4} \right)^3 \times 1000}{64} \right) $$

$$ x = \left( \frac{1.2^3 \times 1000}{64} \right) $$

$$ x = \left( \frac{1.728 \times 1000}{64} \right) $$

$$ x = \left( \frac{1728}{64} \right) $$

$$ x = 27 $$

Therefore, the value of $ x $ is 27.