An electric cable contains a single copper wire of radius 9 mm. Its resistance is $5\Omega$ . This cable is replaced by six insulated copper wires, each of radius 3 mm. The resultant resistance of cable is $\frac{15}{a}$ $\Omega$, where a is prime number. Find the value of 'a'. 

The resistance of the wire is, 

$R = \frac{\rho L}{A}$
$A = \pi r^2$ 
$r = 9 mm$
$A = 81\mathrm{\pi } {\mathrm{mm}}^2$
Therefore,  $R= \frac{\rho L}{ 81\mathrm{\pi }}$

Now, the cable is replaced by six insulated copper wires,
The radius of the copper wires, $r = 3 mm$

So, the new resistance, $R\text{'}= \frac{\rho L}{ 9\mathrm{\pi }}$

$R\text{'} = 9 R$
$We know that, R = 5 \Omega$
$Therefore, R\text{'} = 9 \times 5 \Omega = 45 \Omega$
It is given that; the resultant resistance of cable is $\frac{15}{a} \Omega$
$\frac{15}{a} \Omega = 45 \Omega$

$a = \frac{15}{45} = \frac{1}{3}$