You would need #380 mu"L"# of absolute ethanol to prepare that particular solution.

So, your #"1.2-mL"# sample must contain clarified yeat extract and ethanol.

#V_("solution") = V_("yeast") + V_("ethanol")#

Since the *only source of ethanol* for your solution will be the absolute ethanol, all you need to do is determine how much ethanol your sample contains.

A #"32% v/v"# solution implies that you have #"32 mL"# of ethanol for every #"100 mL"# of solution. This means that your sample will contain

#"1.2 mL solution" * ("32 mL ethanol")/("100 mL solution") = "0.384 mL ethanol"#

Rounded to two sig figs, the answer will be #"0.38 mL"#.

In microliters, this is equal to

#"0.38 mL" * (1000mu"L")/("1 mL") = "380"mu"L"#

As a conclusion, if you mix #"0.38 mL"# absolute ethanol with #"0.82 mL"# clarified yeast extract you'll end up with #"1.2 mL"# of a #"32%"# ethanol solution.