A galvanometer gives full scale deflection with 0 · 006A current....

Given a galvanometer of resistance (of resistance ${R}_{G}$) get converted into a voltmeter of range $0 - 30V$ , when a resistance of ${R}_{S} = 4990Ω$ is connected in series of it and give full scale of deflection with current ${I}_{G} = 0 \cdot 006A$.

Than from equation ${I}_{G} = \dfrac{V}{{R}_{S} + {R}_{G}}$
$0.006 = \dfrac{30}{4990 + G}$ $⇒ {R}_{G} = 10Ω$

Now this galvanometer is connected with a shunt resistance to convert into ammeter of range $0 - 1 \cdot 5A$

Given ${I}_{G} = 0.006$, so current passing shunt will be 
${I}_{S} = I - {I}_{G} = 1 . 5 - 0 \cdot 006 = 1.494A$
Since ${R}_{G}$ and ${R}_{S}$ are in parallel, ${I}_{G}{R}_{G} = {I}_{s}{R}_{S}$
$0.006R = 1.494\left(\dfrac{2n}{249}\right)$ ;   $∵$given ${R}_{S} = \left(\dfrac{2n}{249}\right)$
$∴ n = 5$