An electric dipole has fixed dipole moment $$\vec{p}$$, which makes angle $$\theta$$ with respect to $$x$$-axis. When subjected to an electric field $$\vec{E}_{1} = E\hat{i}$$, it experiences a torque $$\vec{T}_{1} = \tau\hat{k}$$. When subjected to another electric field $$\vec{E}_{2} = \sqrt{3}E_{1}\hat{j}$$, it experiences a torque $$\vec{T}_{2} = -\vec{T}_{1}$$. The angle $$\theta$$ is:

We recall the vector relation for the torque experienced by an electric dipole in a uniform electric field. The formula is stated first:

$$\vec{\tau} \;=\; \vec{p}\;\times\;\vec{E}$$

Here $$\vec{p}$$ is the dipole moment and $$\vec{E}$$ is the electric field. The dipole moment is fixed in the xy-plane and makes an angle $$\theta$$ with the positive $$x$$-axis, so we may write it component-wise as

$$\vec{p}=p\bigl(\cos\theta\,\hat{i}+\sin\theta\,\hat{j}\bigr).$$

Now the first electric field is purely along the $$x$$-axis:

$$\vec{E}_{1}=E\,\hat{i}.$$

Using the cross-product formula, the torque produced by $$\vec{E}_{1}$$ is

$$\vec{\tau}_{1}= \vec{p}\times\vec{E}_{1} = p(\cos\theta\,\hat{i}+\sin\theta\,\hat{j})\times (E\,\hat{i}).$$

Because $$\hat{i}\times\hat{i}=0$$ and $$\hat{j}\times\hat{i}=-\hat{k},$$ only the second term survives:

$$\vec{\tau}_{1}=pE\sin\theta\;(-\hat{k}) = -\,pE\sin\theta\,\hat{k}.$$

The question tells us that for this same field the torque vector is $$\vec{T}_{1}=+\tau\,\hat{k}.$$ Thus the magnitude is clearly

$$\tau_{1}=pE\sin\theta.$$

Next, consider the second electric field, which points along the positive $$y$$-axis and is $$\sqrt{3}$$ times stronger than $$\vec{E}_{1}$$ in magnitude:

$$\vec{E}_{2}=\sqrt{3}\,E\,\hat{j}.$$

The torque produced by this field is obtained in the same way:

$$\vec{\tau}_{2}= \vec{p}\times\vec{E}_{2} = p(\cos\theta\,\hat{i}+\sin\theta\,\hat{j})\times(\sqrt{3}\,E\,\hat{j}).$$

Here $$\hat{j}\times\hat{j}=0$$ and $$\hat{i}\times\hat{j}=+\hat{k},$$ giving

$$\vec{\tau}_{2}=p\sqrt{3}E\cos\theta\,\hat{k}.$$

Therefore the magnitude of the second torque is

$$\tau_{2}=p\sqrt{3}E\cos\theta.$$

The statement in the problem is that the second torque is the negative of the first:

$$\vec{T}_{2}=-\,\vec{T}_{1}.$$

This implies that their magnitudes are equal while their directions are opposite, so we set

$$\tau_{2}=\tau_{1}.$$

Substituting the expressions just obtained, we have

$$p\sqrt{3}E\cos\theta \;=\; pE\sin\theta.$$

The factors $$p$$ and $$E$$ cancel immediately, leaving

$$\sqrt{3}\cos\theta=\sin\theta.$$

Dividing both sides by $$\cos\theta$$ (which is non-zero for the angles listed) gives the tangent:

$$\tan\theta=\sqrt{3}.$$

The standard value that satisfies this equation in the interval $$0^{\circ}\le\theta\le 90^{\circ}$$ is

$$\theta=60^{\circ}.$$

Hence, the correct answer is Option D.