Measured value of R=5 less than
actual value of R i.e., R′=30−(5 of 30)=30−1005×30
R′=28.5Ω ...(i)

Now, let us assume that internal resistance of voltmeter RV. Replacing voltmeter with its internal resistance, we get following circuit.
It is clear that the measured value, R′ should be equal to parallel combination of R and RV. Mathematically,
R′=R+RVR×RV=28.5Ω
Given. R=30Ω
⇒30+RV30RV=28.5
⇒30RV=(28.5×30)+28.5RV
⇒1.5RV=28.5×30
⇒RV=1.528.5×30=19×30
or RV=570Ω