The quadratic hash function for key value $k$ at $i^{th}$ probe sequence is given by
$h(k, i)$ = $(h(k) + c_{1}.i + c_{2}.i^2)$ $mod$ $m$
Where $i = 0, 1, 2, 3.........$ and
$c_{2} ≠ 0$ (since it will degrade to linear probing) so in general we take $c_{2} = 1$ and also
$h(k) = k$ $mod$ $m$
in our case $h(k) = 4594$ $mod$ $100 = 94$
now the value of $h(k , i)$ depends upon $c_{1}$
If we take $c_{1} = 0$ the corresponding probe sequence will be
$h(4594, 0)$ = $(94 + 0•0 + 1•0^2)$ $mod$ 100 = 94 (already occupied)
$h(4594, 1)$ = $(94 + 0•1 + 1•1^2)$ $mod$ 100 = 95 (already occupied)
$h(4594, 2)$ = $(94 + 0•2 + 1•2^2)$ $mod$ 100 = 98 (already occupied)
$h(4594, 3)$ = $(94 + 0•3 + 1• 3^2)$ $mod$ 100 = (103) $mod$ 100 = 3 (which matches our options)
now if we take different value for $c_{1}$ and $c_{2}$ we get different probe sequences
For example if we take $c_{1} = 1$ & $c_{2} = 1$ , the probe sequence we get is 94, 96, 0, 6... (6 not in option)
In wikipedia they have taken quadratic function as $h(k, i)$ = H + $i^{2}$
Where H = k mod m and i = 0, 1, 2, 3,.....
https://en.m.wikipedia.org/wiki/Quadratic_probing
When we take $c_{1} =0$ & $c_{2} = 1$ , we get the same fuction.