An electrical meter of internal resistance 20 Ω gives a full scale deflection when one milliampere current flows through it. The maximum current, that can be measured by using three resistors of resistance 12 Ω each, in milli amperes is :

We know that the equivalent resistance lowers as the resistors are connected in parallel.
By Ohm’s law,
$I=\frac{V}{R}$, where, V is the voltage and R is the resistance of the circuit.
The current in the circuit is inversely proportional to the resistance, as shown by the expression above.
Therefore, when the resistance is at its minimum, the current in the electric metre will be at the maximum level. Connecting the three resistors of can yield a minimum resistance $12 \Omega$ in parallel.
Equivalent resistance of three $12 \Omega$ resistors,
$\frac{1}{R_{eq}}=\frac{1}{12}+\frac{1}{12}+\frac{1}{12}$
$\Rightarrow \frac{1}{R_{eq}}=\frac{3}{12}=\frac{1}{4}$
$\Rightarrow R_{eq}=4 \Omega$
when the internal resistance is, the voltage in the electrical metre $20 \Omega$ is,
$V=I_L{R}_i$ …… (1)

voltage across the resistance $R_{eq}=4 \Omega$ is,
$V=I_2{R}_{eq}$ …… (2)
Since the parallel circuit's voltage does not change, we may say,
$I_L{R}_i=I_2{R}_{eq}$
$\Rightarrow I_2=\frac{I_L{R}_i}{R_{eq}}$
Substitute $1 mA$ for $I_L$, $20 \Omega$ for $R_L$ and $4 \Omega$ for $R_{eq}$ in the above equation.
$I_2=\frac{\left(1 mA\right)\left(20 \Omega \right)}{4 \Omega }$
$\Rightarrow I_2=5 mA$
Thus, The maximum current possible through the electric metre is,
$I_{\max }=1 mA+5 mA$
$\Rightarrow I_{\max }=6 mA$
Hence the correct answer is 6mA.