From the given data, the amount of energy required to break the nucleus of aluminium $$_{13}^{27}$$Al is _________ $$x \times 10^{-3}$$ J
Mass of neutron = 1.00866 u
Mass of proton = 1.00726 u
Mass of Aluminium nucleus = 27.18846 u
(Assume 1 u corresponds to $$x$$ J of energy)
(Round off to the nearest integer)

We begin by noting that the aluminium nucleus in question is $$\,^{27}_{13}\text{Al}\,.$$ Hence, it contains $$Z = 13$$ protons and $$N = A-Z = 27-13 = 14$$ neutrons.

The masses of the individual free nucleons are supplied:

$$m_p = 1.00726\ \text{u}, \qquad m_n = 1.00866\ \text{u}.$$

So, the combined mass of all the separate nucleons is obtained simply by multiplication and addition:

$$\begin{aligned} m_{\text{(13 p)}} &= 13 \times 1.00726 \\[4pt] &= 13.09438\ \text{u},\\[6pt] m_{\text{(14 n)}} &= 14 \times 1.00866 \\[4pt] &= 14.12124\ \text{u}. \end{aligned}$$

Adding these two partial results gives the total mass when the nucleons are unbound:

$$m_{\text{nucleons}} = 13.09438 + 14.12124 = 27.21562\ \text{u}.$$

The measured mass of the bound aluminium nucleus is given as

$$m_{\text{nucleus}} = 27.18846\ \text{u}.$$

The difference between the mass of the free nucleons and the mass of the bound nucleus is called the mass defect:

$$\Delta m = m_{\text{nucleons}} - m_{\text{nucleus}} = 27.21562 - 27.18846 = 0.02716\ \text{u}.$$

The binding energy $$E_b$$ is related to this mass defect through Einstein’s famous relation $$E = mc^2.$$ Since the question explicitly states, “Assume 1 u corresponds to $$x$$ J of energy”, we may write directly

$$E_b = \Delta m \times x \; \text{J}.$$

Numerically substituting the value found for the mass defect, we have

$$E_b = 0.02716\,x \; \text{J}.$$

It is convenient, as asked, to express this result in the form $$\text{(some number)} \times 10^{-3} \; \text{J}.$$ To do so, we simply multiply and divide by $$10^{3}$$:

$$\begin{aligned} E_b &= 0.02716\,x \; \text{J} \\ &= 0.02716\,x \times \frac{10^{3}}{10^{3}} \; \text{J} \\ &= (0.02716 \times 10^{3})\,x \times 10^{-3} \; \text{J} \\ &= 27.16\,x \times 10^{-3} \; \text{J}. \end{aligned}$$

Finally, the instructions tell us to round to the nearest integer. Rounding $$27.16$$ to the nearest whole number gives $$27$$.

Hence, the correct answer is Option 27.