In the given circuit reading of ideal voltmeter inE/2 then internal resistance of cell is

There will be no current flow in the ideal voltmeter. Current in the main circuit is 2E/(3+r). Potential difference across the voltmeter is -(i*r) + (E) = -2Er/(3+r) +E = E/2. {given in question}

       Since along current flow potential decreases, thats why negative sign for i*r.