[tex]\qquad\qquad\huge\underline{{\sf Answer}}[/tex]
Here we go ~
At first, the velocity of sound wave will be :
[tex]\qquad \tt \dashrightarrow \:v = \lambda \cdot f[/tex]
[tex]\qquad \tt \dashrightarrow \:v = 1.76 \times 196[/tex]
Now, the frequency is increased to 784 hertz. but velocity remains same because eave hasn't changed medium
and velocity of wave remains same in a particular medium.
so, we can infer that :
[tex]\qquad \tt \dashrightarrow \: \lambda' = \dfrac{v}{f'} [/tex]
[tex]\qquad \tt \dashrightarrow \: \lambda' = \dfrac{1.76 \times \cancel{196}}{ \cancel{784} \: \: {}^{4} } [/tex]
[tex]\qquad \tt \dashrightarrow \: \lambda' = \dfrac{1.76 }{ {4} } [/tex]
[tex]\qquad \sf \dashrightarrow \: \lambda' = 0.44 \: m[/tex]
so, the new wavelength will be 0.44 m
