How much sr(oh)2 • 8 h2o (m = 265.76) is needed to
prepare 250.0 ml of solution in which [oh–
] = 0.100 m?

First, you need to find the number of moles of OH⁻ in a 250mL solution of 0.100M OH⁻.  To do this, multiply 0.250L by 0.100M to get 0.025mol OH⁻.  Then you use the fact that 1 mole of Sr(OH₂)·8H₂O contains 2 moles of OH⁻ which means that 0.0125mol of Sr(OH)₂·8H₂O contains 0.025mol OH⁻ (0.025/2=0.0125).  Then to find the amount of Sr(OH)₂·8H₂O is needed you multiply its molar mass (265.76g/mol) by 0.0125mol to get 3.322g.
Therefore you need 3.322g of Sr(OH)₂·8H₂O.

I hope that helps.  Let me know if anything is unclear.

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