How can I calculate the molar solubility of BaSO4 given a Ksp of 1.07×10−10?

You always start with the balanced chemical equation, an ICE table, and the ##K_(sp)## expression.

The balanced chemical equation is

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq) I/mol·L⁻¹: -; 0; 0 C/mol·L⁻¹: -; +##x##; +##x## E/mol·L⁻¹: -; ##x##; ##x##

##K_(sp)## = [Ba²⁺][ SO₄²⁻] = ##x × x = x^2## = 1.07 ×10⁻¹⁰

##x = √(1.07 ×10⁻¹⁰)## = 1.03 × 10⁻⁵

Solubility = ##x## mol/L = 1.03 × 10⁻⁵ mol/L

Hope this helps.