$E_{\text{tosses}} = 0.5 \times 1 + 0.5 \times(1+ E_{HaT})$

$\implies E_{\text{tosses}} = 1 + 0.5 E_{HaT}$

$E_{HaT} = 0.25 \times 1 + 0.75 \times (1 + E_{HaT})$

$\implies E_{HaT} = 1 + 0.75 E_{HaT}$

$\implies E_{HaT} = \frac{1}{0.25} = 4$

$\therefore E_{\text{tosses}} = 1 + 0.5 \times 4 = 3$

So, we must toss the coin at least $3$ times to expect a Head.