A sonometer wire of length l vibrates in its fundamental mode when...

The initial fundamental frequency of the sonometer wire is given by:

$f_1 = \dfrac{v}{2l} = 416 \text{ Hz}$

When the length of the wire is doubled ($l' = 2l$), the new fundamental frequency becomes:

$f_1' = \dfrac{v}{2(2l)} = \dfrac{1}{2} \left(\dfrac{v}{2l}\right) = \dfrac{416}{2} = 208 \text{ Hz}$

The natural frequencies of the new wire are integer multiples of its fundamental frequency, i.e., $208 \text{ Hz}, 416 \text{ Hz}, 624 \text{ Hz}, \dots$

Since the wire is excited by the same tuning fork of frequency $416 \text{ Hz}$, the driving frequency matches the second harmonic of the new wire.

In forced oscillations, a body vibrates with the frequency of the driving source. Because the driving frequency corresponds to one of the natural frequencies of the wire, the wire will resonate in its second harmonic mode (forming two loops) and vibrate with a frequency of $416 \text{ Hz}$.