The answer is $ \mathbf {\frac{6}{\pi}}$
Set $ x=\frac{1}{y} $
First equation becomes $$ g(\frac{1}{y}) - \frac{d}{dy}\bigg(\frac{1}{y^2}g'(\frac{1}{y}) \bigg) =0 \,\, \cdots(*)$$
Let $ w= g(\frac{1}{y})$ and hence (*) becomes $$ w(y)+w''(y)=0 \Rightarrow w= c_1\cos y +c_2 \sin y \Rightarrow g(x)= c_1\cos(\frac{1}{x}) +c_2 \sin(\frac{1}{x})$$
Since $\lim_{x\rightarrow \infty} xg(x)=1 $ ,it forces $c_1=0$ and then $ c_2=1$
Now $ g(x)= \sin(\frac{1}{x}) $ and we can easily solve for $a$.