To balance redox equations, I would recommend the half-equation method. This requires that we recognise the equation that describes the oxidation and the reduction:

The initial (unbalanced) simple half equations are:

K₂Cr₂O₇  -->  K⁺  +  Cr³⁺

Ph-CH(OH)-CH₃  -->  Ph-C(=O)-CH₃ or we could simply write C₈H₁₀O   -->   C₈H₈O.... (for the sake if simplicity we will use the second one)

Now for each, balance these by (IN THIS ORDER)

1.  balancing the parent systems (i.e. metal ions and carbon atoms (if required), then
2   balancing O by adding H₂O,
3.  balancing H by adding H⁺ and finally
4.  charges by adding e^-

This creates the two balanced half equations:

K₂Cr₂O₇   +   14 H⁺   +   6 e^-   -->  2 K+  +  2 Cr³⁺ +  7 H₂O

C₈H₁₀O   -->   C₈H₈O  +  2 H⁺   +   2 e^-

Now we need to multiply to equate the number of electrons (in this case we can make them both have 6 e^-)

1 x (K₂Cr₂O₇   +   14 H⁺   +   6 e^-   -->  2 K+  +  2 Cr³⁺ +  7 H₂O )

3 x (C₈H₁₀O   -->   C₈H₈O  +  2 H⁺   +   2 e^- )

Then recombine (by adding them together) the two balanced half equations, cancelling out common terms:

K₂Cr₂O₇   +   3 C₈H₁₀O   + 8 H⁺   -->    3 C₈H₈O   +    2 K⁺   +    2 Cr³⁺    +   7 H₂O

This gives us our balanced redox equation.

10g of phenylethanol (MW = 122g/mol) = 0.0819 moles
1 mole of oxidant oxidises 3 moles of alcohol, so to oxidise 0.0819 moles we need 0.0273 moles of oxidant
0.0273 moles of oxidant = 0.0273 x 294.1816 g = 8.04 g

The alcohol is oxidised to get the ketone. The oxidation state of the C atom changes from 0 to +2.