Given,

Equation of ellipse $\frac{x^2}{a^2}+\frac{{\displaystyle y^2}}{b^2}=1$, eccentricity $e=\frac{\sqrt{3}}{2}$ and length of latusrectum $\frac{2b^2}{a}=1 ...........\left(1\right)$

Now we know that formula of eccentricity $b^2=a^2\left(1-e^2\right)$ so by using this we get $\frac{b^2}{a^2}=1-\frac{3}{4}=\frac{1}{4} ........\left(2\right)$

Now by solving equation $\left(1\right) \& \left(2\right)$ we get,

$a=2 \& b=1$

So sum of major axis and minor axis will be $4+2=6$.