Interpret the figure/data as two dipoles with common center $O$

• Charges $+q$ and $-q$ are separated by $2l$, so they form a dipole of moment $p_1 = q(2l)=2ql.$
• Charges $+2q$ and $-2q$ are separated by $4l$, so they form another dipole of moment $p_2 = 2q(4l)=8ql.$

Since the required potential is given proportional to $\dfrac{ql}{r^2}$, this is the standard dipole potential form, so point $P$ must be far away compared to the size of the charge arrangement.

Use dipole potential formula

The potential due to a dipole at a point on its axial line at distance $r$ (for $r \gg l$) is $V = \frac{1}{4\pi\varepsilon_0}\frac{p}{r^2}.$

If the point lies on the side where the negative end is nearer, the sign is negative: $V = -\frac{1}{4\pi\varepsilon_0}\frac{p}{r^2}.$

Add potentials of the two dipoles

Both dipoles contribute with the same sign as indicated in the question, so total effective dipole moment is $p_{\text{eq}} = p_1+p_2 = 2ql+8ql=10ql.$

Hence, $V = -\frac{1}{4\pi\varepsilon_0}\frac{10ql}{r^2}.$

Substitute the value of Coulomb constant

Using $\frac{1}{4\pi\varepsilon_0}=9\times 10^9,$ we get $V = -9\times 10^9\cdot \frac{10ql}{r^2}$ $V = -90\left(\frac{ql}{r^2}\right)\times 10^9\,\text{V}.$

Compare with the given form

Given, $V=-\alpha\left[\frac{ql}{r^2}\right]\times 10^9\,\text{V}.$

Therefore, $\alpha=90.$

Check against stored answer

The stored correct answer is $27$, but from standard dipole-potential addition the value comes out to be $\boxed{90}.$

So I do not agree with the stored answer.