When a voltmeter put in series, it still reads potential drop and when an ammeter is connected in parallel, it still shows current through it.

Case a

Let $I=$ current through cell, then potential drop read by voltmeter is $V=I·R_V$ (this is reading of voltmeter) Where, $R_V$ is the resistance of voltmeter In loop$A B$, $V_{AB}=I_1\times 1=I_2\times R_A$ and $I=I_1+I_2$

Where, $R_A$ is the resistance of ammeter We substitute for $I_1$ from above equation to get $\Rightarrow I=I_2{R}_A+I_2=I_2\left(R_A+1\right)$

$\Rightarrow I_2=\frac{I}{\left(R_A+1\right)}$

(this is reading of ammeter)

Now given, $\frac{\text{ voltmeter readin'g }}{\text{ ammeter reading }}=1\times {10}^3=\frac{I{R}_V}{\left(\frac{I}{R_A+1}\right)}$

$\mathrm{So},R_V\left(R_A+1\right)=1000$1000$

Case b

Let $I=$ current through cell, then ammeter reading in this case is I.

Also, in loop A B,

$V_{AB}=I_1\times 1=I_2\times R_V$

As, $I=I_1+I_2=I_2{R}_V+I_2$

$=I_2\left(R_V+1\right)$

So, $I_2=\frac{I}{\left(R_V+1\right)}$

Hence, voltmeter reading is $V=I_2{R}_V$ $=\frac{I{R}_V}{\left(R_V+1\right)}$ (this is reading of voltmeter)

Now given, voltmeter reading $÷$ ammeter reading $=0.999\Omega .$

$\mathrm{So},0.999=\frac{\left[\frac{I{R}_V}{\left(R_V+1\right)}\right]}{I}$

$\Rightarrow 0.999=\frac{R_V}{R_V+1}$

$\mathrm{So},R_V=999\Omega$

$\approx {10}^3\Omega$

Substituting $R_V$ in Eq (i), we get

$R_A=\frac{1}{999}$

or $R_A={10}^{-3}\Omega$