By double decomposition method
$\frac{ Eq mass of the metal}{Eq mass of the mettalic nitrate} = \frac{Mass of the metal}{Mass of metallic nitrate}$
$\therefore \frac{E}{E+Eq mass of the NO_3^-} = \frac{10}{18.9}$
$\therefore \frac{E}{E+(14+16 \times 3} = \frac{10}{18.9}$
$\therefore 18.9 E = 10(E+62)$
$\therefore 18.9 E = 10E + 620$
$\therefore 8.9E = 620$
$\therefore E = 69.66$
Atomic Mass x specific heat = 6.4
Atomic mass = $\frac{6.4}{Specific Heat}$
Atomic mass = $\frac{6.4}{0.03} = 213.33$
Now approx atomic = Equivalent mass X Valency
$\therefore$ Valency = Approx Atomic Mass / Equivalent mass
$\therefore Valency = \frac{213.33}{69.66} = 3.06$
Hence Atomic Mass = Equivalent Mass X Valency
Atomic Mass = $69.66 \times 3 = 208.98 U$